In a structure of mixed oxide,oxide ions are | vedclass

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(A) In a $CCP$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$. Number of tetrahedral voids $(TV)$ = $2 	imes 4 = 8$. Number of oc

Vedclass · Accepted Answer

$A_4B_5O_{10}$

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(A) In a $CCP$ structure,the number of oxide ions $(O^{2-})$ per unit cell is $4$. Number of tetrahedral voids $(TV)$ = $2 	imes 4 = 8$. Number of octahedral voids $(OV)$ = $4$. Number of $A^{2+}$ ions = $\frac{1}{5} 	imes 8 = \frac{8}{5}$. Number of $B^{3+}$ ions = $\frac{1}{2} 	imes 4 = 2$. The ratio of $A : B : O$ is $\frac{8}{5} : 2 : 4$. Multiplying the ratio by $5$,we get $8 : 10 : 20$. Simplifying the ratio by dividing by $2$,we get $4 : 5 : 10$. Thus,the formula of the oxide is $A_4B_5O_{10}$.

In a structure of mixed oxide,oxide ions are in $CCP$. One-fifth of tetrahedral voids are occupied by divalent ion $(A^{2+})$ while half of the octahedral voids are occupied by trivalent ion $(B^{3+})$. The formula of the oxide is:

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